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JEE Main 21 January 2026 Shift 1 Question Paper with Answers

21 January 2026 · January session · 72 questions

72 of the 75 questions from the JEE Main 21 January 2026 Shift 1 paper, each with its correct answer and tagged to the chapter it tests. Free to read, no account needed.

3 questions are held back from this paper while we re-check the transcription or the answer key. We would rather show you nothing than show you an answer we are not confident is right.

Physics
25
Chemistry
23
Mathematics
24

Physics — JEE Main 21 January 2026 Shift 1

Q1·PhysicsSingle correct
Potential energy (VVV) versus distance (xxx) is given by the graph, Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.
  1. (A)FBC>FCD>FDE>FABF_{BC} > F_{CD} > F_{DE} > F_{AB}FBC​>FCD​>FDE​>FAB​
  2. (B)FCD>FAB>FBC>FDEF_{CD} > F_{AB} > F_{BC} > F_{DE}FCD​>FAB​>FBC​>FDE​
  3. (C)FCD>FDE>FAB>FBCF_{CD} > F_{DE} > F_{AB} > F_{BC}FCD​>FDE​>FAB​>FBC​
  4. (D)FBC>FAB>FDE>FCDF_{BC} > F_{AB} > F_{DE} > F_{CD}FBC​>FAB​>FDE​>FCD​

Correct answer: (D)

Step-by-step solution →
Q2·PhysicsSingle correct
A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27°C to 87°C. The rate of consumption of the gas is _____ g/s. (Take heat of combustion of gas = 5.0×1045.0 \times 10^{4}5.0×104 J/g) specific heat capacity of water = 4200 J/kg. °C
  1. (A)2.1
  2. (B)4.2
  3. (C)0.42
  4. (D)0.21

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correct
A conducting circular loop of area 1.0 m2\mathrm{m}^{2}m2 is placed perpendicular to a magnetic field which varies as B = sin(100 t) Tesla. If the resistance of the loop is 100 Ω , then the average thermal energy dissipated in the loop in one period is _______J.
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)2π2\pi2π
  3. (C)π\piπ
  4. (D)π2\pi^{2}π2

Correct answer: (C)

Step-by-step solution →
Q4·PhysicsSingle correct
Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at AAA and BBB are 6 cm2\mathrm{cm}^{2}cm2 and 3 cm2\mathrm{cm}^{2}cm2 respectively. The rate of flow will be_____ cm3\mathrm{cm}^{3}cm3/s. (take g = 10 m/s2\mathrm{m/s}^{2}m/s2)
  1. (A)2003\frac{200}{\sqrt{3}}3​200​
  2. (B)2006200\sqrt{6}2006​
  3. (C)2003200\sqrt{3}2003​
  4. (D)1003100\sqrt{3}1003​

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correct
In an experiment the values of two spring constants were measured as k1=(10±0.2)k_{1} = (10 \pm 0.2)k1​=(10±0.2) N/m and k2=(20±0.3)k_{2} = ( 20 \pm 0.3)k2​=(20±0.3) N/m. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :
  1. (A)2.67%
  2. (B)2.33%
  3. (C)1.33%
  4. (D)1.67%

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsSingle correct
A 4 kg mass moves under the influence of a force F⃗=(4t3i^−3tj^)\vec{F} = (4t^{3}\hat{i} - 3t\hat{j})F=(4t3i^−3tj^​) N where t is the time in second. If mass starts from origin at t = 0, the velocity and position after t = 2s will be :
  1. (A)v⃗=3i^+32j^\vec{v} = 3\hat{i} + \frac{3}{2}\hat{j}v=3i^+23​j^​ r⃗=65i^+j^\vec{r} = \frac{6}{5}\hat{i} + \hat{j}r=56​i^+j^​
  2. (B)v⃗=4i^−32j^\vec{v} = 4\hat{i} - \frac{3}{2}\hat{j}v=4i^−23​j^​ r⃗=85i^−j^\vec{r} = \frac{8}{5}\hat{i} - \hat{j}r=58​i^−j^​
  3. (C)v⃗=4i^+52j^\vec{v} = 4\hat{i} + \frac{5}{2}\hat{j}v=4i^+25​j^​ r⃗=85i^+2j^\vec{r} = \frac{8}{5}\hat{i} + 2\hat{j}r=58​i^+2j^​
  4. (D)v⃗=4i^−32j^\vec{v} = 4\hat{i} - \frac{3}{2}\hat{j}v=4i^−23​j^​ r⃗=65i^−j^\vec{r} = \frac{6}{5}\hat{i} - \hat{j}r=56​i^−j^​

Correct answer: (B)

Step-by-step solution →
Q7·PhysicsSingle correct
Consider a modified Bernoulli equation. (P+ABt2)+ρg(h+Bt)+12ρV2=constant\left(P + \frac{A}{Bt^{2}}\right) + \rho g(h + Bt) + \frac{1}{2}\rho V^{2} = \mathrm{constant}(P+Bt2A​)+ρg(h+Bt)+21​ρV2=constant If t has the dimension of time then the dimensions of A and B are ______, ______respectively.
  1. (A)[ML0T−1][ML^{0}T^{-1}][ML0T−1] and [M0LT][M^{0}LT][M0LT]
  2. (B)[ML0T−1][ML^{0}T^{-1}][ML0T−1] and [M0LT−1][M^{0}LT^{-1}][M0LT−1]
  3. (C)[ML0T−2][ML^{0}T^{-2}][ML0T−2] and [M0LT−2][M^{0}LT^{-2}][M0LT−2]
  4. (D)[ML0T−2][ML^{0}T^{-2}][ML0T−2] and [M0LT−1][M^{0}LT^{-1}][M0LT−1]

Correct answer: (B)

Step-by-step solution →
Q8·Physics·Magnetic Field of CurrentSingle correct
A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :
  1. (A)a = g
  2. (B)a > g
  3. (C)a = 0
  4. (D)0 < a < g

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsSingle correct
A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness (13)rd\left(\frac{1}{3}\right)^{\mathrm{rd}}(31​)rd of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :
  1. (A)3KC2K+1\frac{3KC}{2K+1}2K+13KC​
  2. (B)CK2+K\frac{CK}{2+K}2+KCK​
  3. (C)3CK2(2K+1)2\frac{3CK^{2}}{(2K+1)^{2}}(2K+1)23CK2​
  4. (D)4KC3K−1\frac{4KC}{3K-1}3K−14KC​

Correct answer: (A)

Step-by-step solution →
Q10·PhysicsSingle correct
The electric field a plane electromagnetic wave is given by : Ey=69sin⁡[0.6×103x−1.8×1011t]E_{y} = 69 \sin[0.6 \times 10^{3}x - 1.8 \times 10^{11}t]Ey​=69sin[0.6×103x−1.8×1011t] V/m. The expression for magnetic field associated with this electromagnetic wave is_____T.
  1. (A)Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{z} = 2.3 \times 10^{-7} \sin[0.6 \times 10^{3}x - 1.8 \times 10^{11}t]Bz​=2.3×10−7sin[0.6×103x−1.8×1011t]
  2. (B)Bz=2.3×10−7sin⁡[0.6×103x+1.8×1011t]B_{z} = 2.3 \times 10^{-7} \sin[0.6 \times 10^{3}x + 1.8 \times 10^{11}t]Bz​=2.3×10−7sin[0.6×103x+1.8×1011t]
  3. (C)By=69sin⁡[0.6×103x+1.8×1011t]B_{y} = 69 \sin[0.6 \times 10^{3}x + 1.8 \times 10^{11}t]By​=69sin[0.6×103x+1.8×1011t]
  4. (D)By=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{y} = 2.3 \times 10^{-7} \sin[0.6 \times 10^{3}x - 1.8 \times 10^{11}t]By​=2.3×10−7sin[0.6×103x−1.8×1011t]

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correct
In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(= 1.5), the central fringe shifts by 0.2 cm. The value of t is _____ cm.
  1. (A)8×10−48 \times 10^{-4}8×10−4
  2. (B)6.0×10−36.0 \times 10^{-3}6.0×10−3
  3. (C)5.6×10−45.6 \times 10^{-4}5.6×10−4
  4. (D)5.0×10−35.0 \times 10^{-3}5.0×10−3

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correct
A light wave described by E = 60sin⁡(3×1015)t+sin⁡(12×1015)t]60\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t]60sin(3×1015)t+sin(12×1015)t] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) __________ eV. (h = 6.6×10−346.6 \times 10^{-34}6.6×10−34 J-s. and e = 1.6×10−191.6 \times 10^{-19}1.6×10−19C)
  1. (A)5.1
  2. (B)3.8
  3. (C)6.0
  4. (D)7.8

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correct
If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is _________ m. (Atomic number of gold = 79 and 14πϵ0=9×109\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^{9}4πϵ0​1​=9×109 in SI units)
  1. (A)2.95×10−142.95 \times 10^{-14}2.95×10−14
  2. (B)2.95×10−162.95 \times 10^{-16}2.95×10−16
  3. (C)3.85×10−163.85 \times 10^{-16}3.85×10−16
  4. (D)3.85×10−143.85 \times 10^{-14}3.85×10−14

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correct
A uniform rod of mass m and length lll suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ________. (g acceleration due to gravity)
  1. (A)mg/2mg/2mg/2
  2. (B)mg/4mg/4mg/4
  3. (C)mg/3mg/3mg/3
  4. (D)mgmgmg

Correct answer: (B)

Step-by-step solution →
Q15·PhysicsSingle correct
An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30°C. The coefficient of linear expansion of aluminium and steel are 24×10−6/24 \times 10^{-6}/24×10−6/°C and 1.2×10−5/1.2 \times 10^{-5}/1.2×10−5/°C, respectively. The length of this composite rod when its temperature is raised to 100°C, is ______ cm.
  1. (A)120.20
  2. (B)120.15
  3. (C)120.03
  4. (D)120.06

Correct answer: (B)

Step-by-step solution →
Q16·PhysicsSingle correct
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Ω then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _________ N.
  1. (A)7.5×10−27.5 \times 10^{-2}7.5×10−2
  2. (B)5.7×10−35.7 \times 10^{-3}5.7×10−3
  3. (C)5.7×10−25.7 \times 10^{-2}5.7×10−2
  4. (D)7.5×10−37.5 \times 10^{-3}7.5×10−3

Correct answer: (D)

Step-by-step solution →
Q17·PhysicsSingle correct
The given circuit works as :
  1. (A)AND gate
  2. (B)NOR gate
  3. (C)NAND gate
  4. (D)OR gate

Correct answer: (C)

Step-by-step solution →
Q18·PhysicsSingle correct
Two strings (A, B) having linear densities μA=2×10−4\mu_A = 2 \times 10^{-4}μA​=2×10−4 kg/m and μB=4×10−4\mu_B = 4 \times 10^{-4}μB​=4×10−4 kg/m and lengths LA=2.5L_A = 2.5LA​=2.5 m and LB=1.5L_B = 1.5LB​=1.5 m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t1t_1t1​ and t2t_2t2​, respectively, to reach the joint. The ratio t1/t2t_1/t_2t1​/t2​ is :
  1. (A)1.08
  2. (B)1.90
  3. (C)1.67
  4. (D)1.18

Correct answer: (D)

Step-by-step solution →
Q19·PhysicsSingle correct
Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE1.5R_E1.5RE​. This satellite can be moved to a circular orbit of radius 3RE3R_E3RE​ by supplying α×106\alpha \times 10^{6}α×106J of energy. The value of α is ____________. (Take Radius of Earth RE=6×106R_E = 6 \times 10^{6}RE​=6×106 m and g = 10 m/s2^{2}2)
  1. (A)150
  2. (B)500
  3. (C)100
  4. (D)1000

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correct
A point charge of 10−810^{-8}10−8 C is placed at origin. The work done in moving a point charge 2 μC from point A(4, 4, 2) m to B(2, 2, 1) m is ___________J. (14π∈0=9×109(\frac{1}{4\pi \in_0} = 9 \times 10^{9}(4π∈0​1​=9×109 in SI units)
  1. (A)45×10−645 \times 10^{-6}45×10−6
  2. (B)0
  3. (C)30×10−630 \times 10^{-6}30×10−6
  4. (D)15×10−615 \times 10^{-6}15×10−6

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsNumerical
A collimated beam of light of diameter 2 mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is ________ mm.

Correct answer: 280

Step-by-step solution →
Q22·PhysicsNumerical
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of 1 Ω is connected across these points is __________ J.

Correct answer: 1080

Step-by-step solution →
Q23·PhysicsNumerical
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is x2\frac{x}{2}2x​ ML2^{2}2 kg m2^{2}2. The value of x is ___________.

Correct answer: 17

Step-by-step solution →
Q24·PhysicsNumerical
10 mole of oxygen is heated at constant volume from 30°C to 40°C. The change in the internal energy of the gas is _______ cal. (The molecular specific heat of oxygen at constant pressure, CpC_pCp​ = 7 cal./mol °C and R = 2 cal./mol °C.)

Correct answer: 500

Step-by-step solution →
Q25·PhysicsNumerical
In a microscope the objective is having focal length f0f_0f0​ = 2 cm and eye-piece is having focal length fef_efe​ = 4 cm. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ___________.

Correct answer: 100

Step-by-step solution →

Chemistry — JEE Main 21 January 2026 Shift 1

Q26·ChemistrySingle correct
Consider the following reactions. PbCl2+K2CrO4→A+2KClPbCl_2 + K_2CrO_4 \rightarrow A + 2KClPbCl2​+K2​CrO4​→A+2KCl (Hot solution) A+NaOH⇌B+Na2CrO4A + NaOH \rightleftharpoons B + Na_2CrO_4A+NaOH⇌B+Na2​CrO4​ PbSO4+4CH3COONH4→(NH4)2SO4+XPbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + XPbSO4​+4CH3​COONH4​→(NH4​)2​SO4​+X In the above reactions, A, B and X are respectively.
  1. (A)Na2[Pb(OH)2],PbCrO4Na_2[Pb(OH)_2], PbCrO_4Na2​[Pb(OH)2​],PbCrO4​ and (NH4)2[Pb(CH3COO)4](NH_4)_2[Pb(CH_3COO)_4](NH4​)2​[Pb(CH3​COO)4​]
  2. (B)PbCrO4PbCrO_4PbCrO4​, Na2[Pb(OH)4]Na_2[Pb(OH)_4]Na2​[Pb(OH)4​] and [Pb(NH3)4]SO4[Pb(NH_3)_4]SO_4[Pb(NH3​)4​]SO4​
  3. (C)Na2[Pb(OH)2]Na_2[Pb(OH)_2]Na2​[Pb(OH)2​], PbCrO4PbCrO_4PbCrO4​ and [Pb(NH3)4]SO4[Pb(NH_3)_4]SO_4[Pb(NH3​)4​]SO4​
  4. (D)PbCrO4PbCrO_4PbCrO4​, Na2[Pb(OH)4]Na_2[Pb(OH)_4]Na2​[Pb(OH)4​] and (NH4)2[Pb(CH3COO)4](NH_4)_2[Pb(CH_3COO)_4](NH4​)2​[Pb(CH3​COO)4​]

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correct
Which of the following represents the correct trend for the mentioned property ? A. F > P > S > B – Frist Ionization Energy B. Cl > F > S > P – Electron Affinity C. K > Al > Mg > B – Metallic character D. K2O>Na2O>MgO>Al2O3K_2O > Na_2O > MgO > Al_2O_3K2​O>Na2​O>MgO>Al2​O3​ – Basic character Choose the correct answer from the option given below.
  1. (A)A, B and D only
  2. (B)A, B, C and D
  3. (C)A and B only
  4. (D)B and C only

Correct answer: (A)

Step-by-step solution →
Q28·ChemistrySingle correct
Identify A in the following reaction.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q29·ChemistrySingle correct
A hydrocarbon 'P' (C4H8C_4H_8C4​H8​) on reaction with HCl gives an optically active compound 'Q' (C4H9ClC_4H_9ClC4​H9​Cl) which on reaction with one mole of ammonia gives compound 'R' (C4H11NC_4H_{11}NC4​H11​N). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q30·ChemistrySingle correct
Given below are two statements : Statement I : The number of pairs among [SiO2,CO2][SiO_2, CO_2][SiO2​,CO2​], [SnO,SnO2][SnO, SnO_2][SnO,SnO2​], [PbO,PbO2][PbO, PbO_2][PbO,PbO2​] and [GeO,GeO2][GeO, GeO_2][GeO,GeO2​], which contain oxides that are both amphoteric is 2. Statement II : BF3BF_3BF3​ is an electron deficient molecule can act as a lewis acid, forms adduct with NH3NH_3NH3​ and has a trigonal planar geometry. In the light of the above statement, choose the correct answer from the option given below.
  1. (A)Both Statement I and Statement II are true.
  2. (B)Both Statement I and Statement II are false.
  3. (C)Statement I is true but Statement II is false.
  4. (D)Statement I is false Statement II is true.

Correct answer: (A)

Step-by-step solution →
Q31·ChemistrySingle correct
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :
  1. (A)C2H4C_2H_4C2​H4​
  2. (B)C4H10C_4H_{10}C4​H10​
  3. (C)C2H2C_2H_2C2​H2​
  4. (D)C2H6C_2H_6C2​H6​

Correct answer: (C)

Step-by-step solution →
Q32·ChemistrySingle correct
14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. Which of the following statements is incorrect ? [Given : Molar mass in g mol−1mol^{-1}mol−1 of Ca–40, Cl–35.5,H–1]
  1. (A)0.35 mol of H2H_2H2​ gas is evolved.
  2. (B)7.84 L of H2H_2H2​ gas is evolved.
  3. (C)33.3 of CaCl2CaCl_2CaCl2​ is produced.
  4. (D)The limiting reagent is calcium metal.

Correct answer: (C)

Step-by-step solution →
Q33·ChemistrySingle correct
In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g mol−1mol^{-1}mol−1). Molar mass of barium sulphate is 233 g mol−1mol^{-1}mol−1.
  1. (A)4.55%
  2. (B)10.30%
  3. (C)21.97%
  4. (D)16.48%

Correct answer: (C)

Step-by-step solution →
Q34·ChemistrySingle correct
Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ2PQ_2PQ2​. When 1 g of PQ is dissolved in 50 g of solvent 'A'. ΔTb\Delta T_bΔTb​ was 1.176 K while when 1 g of PQ2PQ_2PQ2​ is dissolved in 50 g of solvent 'A', ΔTb\Delta T_bΔTb​ was 0.689 K. (KbK_bKb​ of 'A' = 5 K kg mol−1mol^{-1}mol−1). The molar masses of elements P and Q (in g mol−1mol^{-1}mol−1) respectively, are :
  1. (A)70, 110
  2. (B)65, 145
  3. (C)60, 25
  4. (D)25, 60

Correct answer: (D)

Step-by-step solution →
Q35·ChemistrySingle correct
An organic compound "P" of molecular formula C6H12O3C_6H_{12}O_3C6​H12​O3​ gives positive lodoform test but negative Tollen's test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollen's test and also iodoform test. The structure of "P" is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (B)

Step-by-step solution →
Q36·Chemistry·Electronic Effects and StabilitySingle correct
From the following, the least stable structure is :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (C)

Step-by-step solution →
Q37·ChemistrySingle correct
MnO42−MnO_4^{2-}MnO42−​ , in acidic medium, disproportionates to :
  1. (A)Mn2O7Mn_2O_7Mn2​O7​ and MnO2MnO_2MnO2​
  2. (B)MnO4−MnO_4^-MnO4−​ and MnO
  3. (C)MnO4−MnO_4^-MnO4−​ and MnO2MnO_2MnO2​
  4. (D)Mn2O7Mn_2O_7Mn2​O7​ and MnO

Correct answer: (C)

Step-by-step solution →
Q38·ChemistrySingle correct
Given below are two statements: Statement I: The number of species among SF4_44​, NH4+_4^+4+​, [NiCl4_44​], XeF4_44​, [PtC14_44​]2−^{2-}2−, SeF4_44​ and [Ni(CN)4_44​]2−^{2-}2−, that have tetrahedral geometry is 3. Statement II: In the set [NO2_22​, BeH2_22​, BF3_33​, AIC13_33​], all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II are true

Correct answer: (C)

Step-by-step solution →
Q39·Chemistry·IsomerismSingle correct
Identify correct statement from the following : A. Propanal and propanone are functional isomers. B. Ethoxyethane and methoxypropane are metamers. C. But-2-ene shows optical isomerism. D. But-1-ene and but-2-ene are functional isomers. E. Pentane and 2, 2-dimethyl propane are chain isomers. Choose the correct answer from the options given below :
  1. (A)B, C and D only
  2. (B)A, B and C only
  3. (C)A, B and E only
  4. (D)C, D and E only

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correct
Identify the correct statements. A. Arginine and Tryptophan are essential amino acids. B. Histidine does not contain heterocyclic ring in its structure. C. Proline is a six membered cyclic ring amino acid. D. Glycine does not have chiral centre. E. Cysteine has characteristic feature of side chain as MeS-CH2_22​-CH2_22​-. Choose the correct answer from the options given below: Option
  1. (A)C and E Only
  2. (B)B and E Only
  3. (C)C and D Only
  4. (D)A and D Only

Correct answer: (D)

Step-by-step solution →
Q41·ChemistrySingle correct
Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q42·ChemistrySingle correct
For the reaction, N2_22​O4_44​ ⇌ 2NO2_22​, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is –5.40 kJ mol−1^{-1}−1. B. As AG⊖^{\ominus}⊖ in graph is positive, N2_22​O4_44​ will not dissociate into NO2_22​ at all. C. Reverse reaction will go to completion. D. When 1 mole of N2_22​O4_44​ changes into equilibrium mixture, value of ΔG⊖^{\ominus}⊖ = –0.84 kJ mol−1^{-1}−1 E. When 2 mole of NO2_22​, changes into equilibrium mixture, ΔG⊖^{\ominus}⊖ for equilibrium mixture is –6.24 kJ mol−1^{-1}−1. E. When 2 mole of NO2_22​, changes into equilibrium mixture, ΔG⊖^{\ominus}⊖ for equilibrium mixture is –6.24 kJ mol−1^{-1}−1. Choose the correct answer from the options given below :
  1. (A)D and E only
  2. (B)C and E only
  3. (C)A and D only
  4. (D)B and C only

Correct answer: (A)

Step-by-step solution →
Q43·ChemistrySingle correct
Given below are two statements: Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies. Statement II: The frequency of second line of Balmer series obtained from He+^++ is equal to that of first line of Lyman series obtained from hydrogen atom. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is false but Statement II is true
  4. (D)Statement I is true but Statement II is false

Correct answer: (A)

Step-by-step solution →
Q44·ChemistryNumerical
Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol−1^{-1}−1. If k1_11​ and k2_22​ are the rate constants of first and second reaction respectively at 300 K, then In k2k1\frac{k_2}{k_1}k1​k2​​ will be …….. . (nearest integer) [R=8.3 J K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 8

Step-by-step solution →
Q45·ChemistryNumerical
The pH and conductance of a weak acid (HX) was found to be 5 and 4 × 10−5^{-5}−5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm2^22 were at a distance of 15 cm apart. The value of the limiting molar conductivity is ……….. S m2^22 mol−1^{-1}−1. (nearest integer) (Given: degree of dissociation of the weak acid (a) << 1)

Correct answer: 6

Step-by-step solution →
Q46·ChemistryNumerical
Use the following data : One mole each of A2_22​(g) and B2_22​(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500K. A2_22​(g) + B2_22​(g) ⇌ 2AB(g) The value of x (in kJ mol−1^{-1}−1) is ……….. (Nearest integer) (Given: log K=2.2 R=8.3 JK−1^{-1}−1 mol−1^{-1}−1)
SubstanceΔf\Delta_fΔf​H⊖^{\ominus}⊖(500K) / kJ mol−1^{-1}−1S⊖^{\ominus}⊖(500K) / J K−1^{-1}−1 mol−1^{-1}−1
AB(g)32222
A2_22​(g)6146
B2_22​(g)X280

Correct answer: 70

Step-by-step solution →
Q47·ChemistryNumerical
Consider the following reaction sequence The percentage of nitrogen in product ‘T’ formed is ________%. (Nearest integer) (Given molar mass in g mol−1^{-1}−1 H:1, C:12, N:14, O:16)

Correct answer: 20

Step-by-step solution →
Q48·ChemistryNumerical
Consider the following reactions: NaCl+K2_22​Cr2_22​O7_77​+H2_22​SO4_44​→A+KHSO4_44​+NaHSO4_44​+H2_22​O A + NaOH → B + NaCl + H2_22​O B + H2_22​SO4_44​ + H2_22​O2_22​ → C + Na2_22​SO4_44​ + H2_22​O In the product 'C', 'X' is the number of O22−_2^{2-}22−​ units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X + Y + Z is ________ .

Correct answer: 13

Step-by-step solution →

Mathematics — JEE Main 21 January 2026 Shift 1

Q49·MathematicsSingle correct
If the domain of the function f(x)=cos⁡−1(2x−511−3x)+sin⁡−1(2x2−3x+1)f(x) = \cos^{-1}\left(\frac{2x-5}{11-3x}\right) + \sin^{-1}(2x^{2} - 3x + 1)f(x)=cos−1(11−3x2x−5​)+sin−1(2x2−3x+1) is the interval [α,β][\alpha, \beta][α,β], then α+2β\alpha + 2\betaα+2β is equal to :
  1. (A)1
  2. (B)3
  3. (C)5
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q50·MathematicsSingle correct
The area of the region, inside the ellipse x2+4y2=4x^{2} + 4y^{2} = 4x2+4y2=4 and outside the region bounded by the curves y=∣x∣−1y = |x| - 1y=∣x∣−1 and y=1−∣x∣y = 1 - |x|y=1−∣x∣, is :
  1. (A)2(π−1)2(\pi - 1)2(π−1)
  2. (B)2π−122\pi - \frac{1}{2}2π−21​
  3. (C)3(π−1)3(\pi - 1)3(π−1)
  4. (D)2π−12\pi - 12π−1

Correct answer: (A)

Step-by-step solution →
Q51·MathematicsSingle correct
The number of relations, defined on the set {a,b,c,d}\{a, b, c, d\}{a,b,c,d}, which are both reflexive and symmetric, is equal to:
  1. (A)256
  2. (B)16
  3. (C)1024
  4. (D)64

Correct answer: (D)

Step-by-step solution →
Q52·MathematicsSingle correct
Let a point A lie between the parallel lines L1L_{1}L1​ and L2L_{2}L2​ such that its distances from L1L_{1}L1​ and L2L_{2}L2​ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L1L_{1}L1​ and L2L_{2}L2​ respectively, is :
  1. (A)15615\sqrt{6}156​
  2. (B)27
  3. (C)21321\sqrt{3}213​
  4. (D)12212\sqrt{2}122​

Correct answer: (C)

Step-by-step solution →
Q53·MathematicsSingle correct
Let a⃗=−i^+2j^+2k^\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}a=−i^+2j^​+2k^, b⃗=8i^+7j^−3k^\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}b=8i^+7j^​−3k^ and c⃗\vec{c}c be a vector such that a⃗×c⃗=b⃗\vec{a} \times \vec{c} = \vec{b}a×c=b. If c⃗.(i^+j^+k^)=4\vec{c}.(\hat{i} + \hat{j} + \hat{k}) = 4c.(i^+j^​+k^)=4, then ∣a⃗+c⃗∣2|\vec{a} + \vec{c}|^{2}∣a+c∣2 is equal to :
  1. (A)33
  2. (B)30
  3. (C)35
  4. (D)27

Correct answer: (D)

Step-by-step solution →
Q54·MathematicsSingle correct
Let a1a_{1}a1​, a2a_{2}a2​, a3a_{3}a3​,….. be a G.P. of increasing positive terms such that a2.a3.a4=64a_{2}.a_{3}.a_{4} = 64a2​.a3​.a4​=64 and a1+a3+a5=8137a_{1} + a_{3} + a_{5} = \frac{813}{7}a1​+a3​+a5​=7813​ . Then a3+a5+a7a_{3} + a_{5} + a_{7}a3​+a5​+a7​ is equal to :
  1. (A)3256
  2. (B)3252
  3. (C)3244
  4. (D)3248

Correct answer: (B)

Step-by-step solution →
Q55·MathematicsSingle correct
Let c⃗\vec{c}c and d⃗\vec{d}d be vectors such that ∣c⃗+d⃗∣=29|\vec{c} + \vec{d}| = \sqrt{29}∣c+d∣=29​ and c⃗×(2i^+3j^+4k^)=(2i^+3j^+4k^)×d⃗\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}c×(2i^+3j^​+4k^)=(2i^+3j^​+4k^)×d. If λ1,λ2(λ1>λ2)\lambda_{1}, \lambda_{2}(\lambda_{1} > \lambda_{2})λ1​,λ2​(λ1​>λ2​) are the possible values of (c⃗+d⃗).(−7i^+2j^+3k^)(\vec{c} + \vec{d}).(-7\hat{i} + 2\hat{j} + 3\hat{k})(c+d).(−7i^+2j^​+3k^), then the equation K2x2+(K2−5K+λ1)xy+(3K+λ22)y2−8x+12y+λ2=0K^{2}x^{2} + (K^{2} - 5K + \lambda_{1})xy + \left(3K + \frac{\lambda_{2}}{2}\right)y^{2} - 8x + 12y + \lambda_{2} = 0K2x2+(K2−5K+λ1​)xy+(3K+2λ2​​)y2−8x+12y+λ2​=0 represents a circle, for k equal to :
  1. (A)4
  2. (B)1
  3. (C)–1
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q56·MathematicsSingle correct
Let y=y(x)y = y(x)y=y(x) be the solution curve of the differential equation (1+x2)dy+(y−tan⁡−1x)dx=0(1 + x^{2})dy + (y - \tan^{-1}x)dx = 0(1+x2)dy+(y−tan−1x)dx=0, y(0)=1y(0) = 1y(0)=1. Then the value of y(1)y(1)y(1) is :
  1. (A)2eπ4+π4−1\frac{2}{e^{\frac{\pi}{4}}} + \frac{\pi}{4} - 1e4π​2​+4π​−1
  2. (B)2eπ4−π4−1\frac{2}{e^{\frac{\pi}{4}}} - \frac{\pi}{4} - 1e4π​2​−4π​−1
  3. (C)4eπ4+π2−1\frac{4}{e^{\frac{\pi}{4}}} + \frac{\pi}{2} - 1e4π​4​+2π​−1
  4. (D)4eπ4−π2−1\frac{4}{e^{\frac{\pi}{4}}} - \frac{\pi}{2} - 1e4π​4​−2π​−1

Correct answer: (A)

Step-by-step solution →
Q57·MathematicsSingle correct
The number of strictly increasing functions f from the set {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}{1,2,3,4,5,6} to the set (1,2,3,…,9)(1, 2, 3,…,9)(1,2,3,…,9) such that f(i)≠if(i) \neq if(i)=i for 1≤i≤61 \leq i \leq 61≤i≤6, is equal to :
  1. (A)21
  2. (B)27
  3. (C)22
  4. (D)28

Correct answer: (D)

Step-by-step solution →
Q58·MathematicsSingle correct
Let f:R→(0,∞)f : R \to (0, \infty)f:R→(0,∞) be a twice differentiable function such that f(3)=18f(3) = 18f(3)=18, f′(3)=0f'(3) = 0f′(3)=0 and f′′(3)=4f''(3) = 4f′′(3)=4. Then lim⁡x→1(log⁡e(f(2+x)f(3))18(x−1)2)\lim_{x \to 1}\left( \log_{e}\left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)limx→1​(loge​(f(3)f(2+x)​)(x−1)218​) is equal to :
  1. (A)1
  2. (B)9
  3. (C)2
  4. (D)18

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsSingle correct
Let the foci of hyperbola coincide with the foci of the ellipse x236+y216=1\frac{x^{2}}{36} + \frac{y^{2}}{16} = 136x2​+16y2​=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:
  1. (A)12
  2. (B)16
  3. (C)965\frac{96}{\sqrt{5}}5​96​
  4. (D)24524\sqrt{5}245​

Correct answer: (C)

Step-by-step solution →
Q60·MathematicsSingle correct
The value of ∫−π/6π/6(π+4x111−sin⁡(∣x∣+π/6))dx\int_{-\pi/6}^{\pi/6}\left( \frac{\pi + 4x^{11}}{1 - \sin(|x| + \pi/6)} \right) dx∫−π/6π/6​(1−sin(∣x∣+π/6)π+4x11​)dx is equal to
  1. (A)2π2\pi2π
  2. (B)4π4\pi4π
  3. (C)8π8\pi8π
  4. (D)6π6\pi6π

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correct
Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, x>yx > yx>y, be 8 and 16 respectively. Two numbers are chosen from {1,2,3,x–4,y,5}\{1, 2, 3, x–4, y, 5\}{1,2,3,x–4,y,5} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:
  1. (A)35\frac{3}{5}53​
  2. (B)45\frac{4}{5}54​
  3. (C)25\frac{2}{5}52​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correct
Let (α, β, γ) be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line r⃗=(−i^+3j^+k^)+λ(2i^+3j^−k^)\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})r=(−i^+3j^​+k^)+λ(2i^+3j^​−k^). Then the length of the projection of the vector αi^+βj^+γk^\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}αi^+βj^​+γk^ on the vector 6i^+2j^+3k^6\hat{i}+2\hat{j}+3\hat{k}6i^+2j^​+3k^ is :
  1. (A)157\frac{15}{7}715​
  2. (B)4
  3. (C)187\frac{18}{7}718​
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q63·MathematicsSingle correct
Let PQ and MN be two straight lines touching the circle x2+y2−4x−6y−3=0x^{2}+y^{2}-4x-6y-3=0x2+y2−4x−6y−3=0 at the points A and B respectively. Let O be the centre of the circle and ∠AOB = π/3. Then the locus of the point of intersection of the lines PQ and MN is:
  1. (A)3(x2+y2)−18x−12y+25=03(x^{2}+y^{2})-18x-12y+25=03(x2+y2)−18x−12y+25=0
  2. (B)x2+y2−12x−18y−25=0x^{2}+y^{2}-12x-18y-25=0x2+y2−12x−18y−25=0
  3. (C)x2+y2−18x−12y−25=0x^{2}+y^{2}-18x-12y-25=0x2+y2−18x−12y−25=0
  4. (D)3(x2+y2)−12x−18y−25=03(x^{2}+y^{2})-12x-18y-25=03(x2+y2)−12x−18y−25=0

Correct answer: (D)

Step-by-step solution →
Q64·MathematicsSingle correct
If the coefficient of x in the expansion of (ax2+bx+c)(1−2x)26(ax^{2}+bx+c)(1-2x)^{26}(ax2+bx+c)(1−2x)26 is −56-56−56 and the coefficients of x2x^{2}x2 and x3x^{3}x3 are both zero, then a + b + c is equal to
  1. (A)1300
  2. (B)1500
  3. (C)1403
  4. (D)1483

Correct answer: (C)

Step-by-step solution →
Q65·MathematicsSingle correct
If x2+x+1=0x^{2}+x+1=0x2+x+1=0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+...+(x25+1x25)4\left(x+\frac{1}{x}\right)^{4}+\left(x^{2}+\frac{1}{x^{2}}\right)^{4}+\left(x^{3}+\frac{1}{x^{3}}\right)^{4}+...+\left(x^{25}+\frac{1}{x^{25}}\right)^{4}(x+x1​)4+(x2+x21​)4+(x3+x31​)4+...+(x25+x251​)4 is :
  1. (A)128
  2. (B)162
  3. (C)175
  4. (D)145

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correct
The value of cosec⁡10∘−3 sec⁡10∘\operatorname{cosec}10^{\circ}-\sqrt{3}\ \sec10^{\circ}cosec10∘−3​ sec10∘ is equal to:
  1. (A)4
  2. (B)2
  3. (C)8
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correct
The sum of all the roots of the equation (x−1)2−5∣x−1∣+6=0(x-1)^{2}-5|x-1|+6=0(x−1)2−5∣x−1∣+6=0, is:
  1. (A)4
  2. (B)3
  3. (C)1
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q68·MathematicsNumerical
Let f : R→R be a twice differentiable function such that the quadratic equation f(x)m2−2f′(x)m+f′′(x)=0f(x)m^{2}-2f'(x)m+f''(x)=0f(x)m2−2f′(x)m+f′′(x)=0 in m, has two equal roots for every x∈Rx\in Rx∈R. If f(0)=1f(0)=1f(0)=1, f′(0)=2f'(0)=2f′(0)=2 and (α, β) is the largest interval in which the function f(log⁡ex−x)f(\log_{e}x-x)f(loge​x−x) is increasing, then α + β is equal to

Correct answer: 1

Step-by-step solution →
Q69·MathematicsNumerical
Let a1=1a_{1}=1a1​=1and for n≥1n\geq 1n≥1, an+1=12an+n2−2n−1n2(n+1)2a_{n+1}=\frac{1}{2}a_{n}+\frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}an+1​=21​an​+n2(n+1)2n2−2n−1​. Then ∣∑n=1∞(an−2n2)∣\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|​∑n=1∞​(an​−n22​)​ is equal to ________.

Correct answer: 2

Step-by-step solution →
Q70·MathematicsNumerical
Let S={(m, n): m, n ∈ {1, 2, 3, ....., 50}}. If the number of elements (m, n) in S such that 6m+9n6^{m}+9^{n}6m+9n is a multiple of 5 is p and the number of elements (m, n) in S such that m + n is a square of a prime number is q, then p + q is equal to……..

Correct answer: 1333

Step-by-step solution →
Q71·MathematicsNumerical
For some α, β ∈ R, let A=[α212]A=\begin{bmatrix}\alpha & 2\\ 1 & 2\end{bmatrix}A=[α1​22​] and B=[111β]B=\begin{bmatrix}1 & 1\\ 1 & \beta\end{bmatrix}B=[11​1β​] be such that A2−4A+2I=B2−3B+I=OA^{2}-4A+2I=B^{2}-3B+I=OA2−4A+2I=B2−3B+I=O. Then (det⁡(adj⁡(A3−B3)))2(\det(\operatorname{adj}(A^{3}-B^{3})))^{2}(det(adj(A3−B3)))2 is equal to ………

Correct answer: 225

Step-by-step solution →
Q72·MathematicsNumerical
6∫0π∣(sin⁡3x+sin⁡2x+sin⁡x)∣dx6\int_{0}^{\pi}\left|(\sin 3x+\sin 2x+\sin x)\right|dx6∫0π​∣(sin3x+sin2x+sinx)∣dx is equal to……..

Correct answer: 17

Step-by-step solution →

Chapters tested in this paper

Open any chapter to practise every past-year question on that topic across all papers, and see how often it has actually appeared.

  • Properties of Solids and Liquids 172/186
  • Three Dimensional Geometry 176/186
  • Matrices and Determinants 180/186
  • Sets, Relations and Functions 165/186
  • Current Electricity 160/186
  • Sequence and Series 164/186
  • p-Block Elements 164/186
  • Definite Integration 168/186
  • Rotational Motion 172/186
  • Redox Reactions and Electrochemistry 177/186
  • Geometrical Optics 172/186
  • Vector Algebra 173/186
  • Differential Equations 167/186
  • Probability 176/186
  • Permutations and Combinations 162/186
  • Magnetic Field of Current 147/186
  • Binomial Theorem and Its Simple Applications 158/186
  • Electromagnetic Waves 144/186
  • Chemical Bonding and Molecular Structure 151/186
  • Application of Derivatives 139/186
  • Limits and Continuity 149/186
  • d- and f-Block Elements 126/186
  • Thermodynamics 154/186
  • Aldehydes and Ketones 135/186
  • Solutions 158/186
  • Laws of Motion 130/186
  • Chemical Thermodynamics 165/186
  • Units and Measurements 149/186
  • Hydrocarbons 126/186
  • Complex Numbers 165/186
  • Trigonometric Functions 144/186
  • Biomolecules 162/186
  • Chemical Kinetics 169/186
  • Gravitation 152/186
  • Atomic Structure 161/186
  • Electronic Devices 145/186
  • Circles 142/186
  • Dual Nature of Matter and Radiation 155/186
  • Amines 133/186
  • Quadratic Equations 148/186
  • Work, Energy and Power 132/186
  • Some Basic Concepts in Chemistry 129/186
  • Electromagnetic Induction 120/186
  • Wave Optics 130/186
  • Area Under Curves 139/186
  • Purification and Characterisation of Organic Compounds 116/186
  • Straight Lines 114/186
  • Waves 109/186
  • Capacitors and Dielectrics 115/186
  • Atoms 112/186
  • Classification of Elements and Periodicity in Properties 113/186
  • Inverse Trigonometric Functions 93/186
  • Electronic Effects and Stability 74/186
  • Hyperbola 77/186
  • Electric Potential 63/186
  • Isomerism 51/186
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